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A nonconvex gradient energy can lose weak lower semicontinuity

Statement refuted

Counterexample. Assume the Axiom of Choice, the ultrafilter lemma, DC and HB (The Axiom of Choice, The ultrafilter extension principle (UL/BPI), The axiom of dependent choice: a relation in which every element is related to something admits an $N$ -indexed chain, The real dominated-extension principle as an additional hypothesis over ZF). On $Ω = (0, 1)$ let $w : R \to R$ be the $1$ -periodic function with $w (t) = t$ on $[0, \frac{1}{2}]$ and $w (t) = 1 - t$ on $[\frac{1}{2}, 1]$ , and put $u_{k} (x) = \frac{1}{k} w (k x)$ for integers $k \geq 1$ . Then $u_{k} \to 0$ uniformly, $u_{k}^{'} \in {\pm 1}$ almost everywhere, and $u_{k} ⇀ 0$ in $H^{1} (0, 1)$ . For the nonnegative integrand $f (ξ) = (ξ^{2} - 1)^{2}$ , which is not convex since $f (0) = 1 > \frac{1}{2} f (- 1) + \frac{1}{2} f (1) = 0$ , The integral is interpreted in $[0, + \infty]$ on $H^{1} (0, 1)$ ; it need not be finite for every $H^{1}$ function. $I (u) = \int_{0}^{1} (u^{'} (x)^{2} - 1)^{2} d x$ satisfies $I (u_{k}) = 0$ for every $k$ , while $I (0) = \int_{0}^{1} 1 d x = 1$ . Hence $I (0) > \lim_{k} I (u_{k}) = 0$ and $I$ is not weakly sequentially lower semicontinuous. Also $I$ is not convex: $I (u_{1}) = I (- u_{1}) = 0 < I ((u_{1} - u_{1}) / 2) = 1$ , so A convex norm-lower-semicontinuous functional is weakly lower semicontinuous does not apply. Nevertheless $I$ attains its minimum $0$ at every $u_{k}$ . This example does not refute the existence conclusion of The direct method for convex integral functionals with convexity removed: it also lies outside that theorem's $n \geq 2$ and $p = 2$ upper-growth hypotheses.

Facts & Assumptions

Given: The Axiom of Choice (for ACL), the ultrafilter lemma, DC and HB; the interval $Ω = (0, 1)$ ; the $1$ -periodic continuous function $w$ with $w (t) = t$ on $[0, \frac{1}{2}]$ and $w (t) = 1 - t$ on $[\frac{1}{2}, 1]$ ; the functions $u_{k} (x) = \frac{1}{k} w (k x)$ ; and the integrand $f (ξ) = (ξ^{2} - 1)^{2}$ with $I (u) = \int_{0}^{1} (u^{'} (x)^{2} - 1)^{2} d x$ . The weak compactness step uses the ultrafilter lemma, DC and HB (The ultrafilter extension principle (UL/BPI), The axiom of dependent choice: a relation in which every element is related to something admits an $N$ -indexed chain, The real dominated-extension principle as an additional hypothesis over ZF).

[F1]

The function $w$ is continuous, $1$ -periodic, satisfies $∣ w ∣ \leq \frac{1}{2}$ and $w (0) = w (1) = 0$ , and is piecewise linear with $∣ w^{'} ∣ = 1$ almost everywhere. Hence $u_{k} (x) = \frac{1}{k} w (k x)$ is absolutely continuous on $[0, 1]$ with a.e. derivative $u_{k}^{'} (x) = w^{'} (k x) \in {\pm 1}$ ; by the ACL characterisation $u_{k} \in W^{1, 2} (0, 1)$ with weak derivative $u_{k}^{'}$ (The ACL characterisation of $W^{1, p}$ , Integer-order Sobolev spaces and their norms).

[F2]

For $φ \in L^{2} (0, 1)$ the maps $u \mapsto \int_{0}^{1} u φ$ and $u \mapsto \int_{0}^{1} u^{'} φ$ are bounded linear functionals on $W^{1, 2} (0, 1)$ with operator norm at most $∥ φ ∥_{L^{2}}$ , by Cauchy-Schwarz against the two components of the Sobolev norm (Integer-order Sobolev spaces and their norms).

[F3]

$W^{1, 2} (0, 1)$ is a reflexive Banach space, so every norm-bounded sequence in it has a weakly convergent subsequence under the ultrafilter lemma, DC and HB (W^{1,p}(Omega) is reflexive for 1<p<infinity, Reflexivity is equivalent to weak subsequential compactness of bounded sequences).

[F4]

If $y \in L^{2} (0, 1; R)$ has $\int y φ = 0$ for all real $φ \in L^{2}$ , take $φ = y$ to obtain $∥ y ∥_{2}^{2} = 0$ , so $y = 0$ as an $L^{2}$ class. This directly identifies the subsequential Sobolev limit; uniqueness of weak probability-measure limits is not used.

[F5]

The weak lower semicontinuity lemma assumes convexity of the functional (A convex norm-lower-semicontinuous functional is weakly lower semicontinuous). The convex integral existence theorem also assumes $n \geq 2$ and a $p$ -growth upper bound (The direct method for convex integral functionals); the present interval and quartic integrand fail these hypotheses for $p = 2$ . Nonconvexity is checked by the midpoint inequality (Convex and strictly convex functionals on a convex subset of a real vector space), and existence here is decided by the explicit values of $I$ .

Counterexample

technique · direct computation along the oscillating sawtooth sequence

1.1F1algebra

The sequence and its bounds. By [F1] the functions $u_{k}$ lie in $W^{1, 2} (0, 1)$ with $∣ u_{k} ∣ \leq \frac{1}{2 k}$ and $∣ u_{k}^{'} ∣ = 1$ almost everywhere; hence $∥ u_{k} ∥_{L^{2}} \leq \frac{1}{2 k} \to 0$ and $∥ u_{k}^{'} ∥_{L^{2}} = 1$ for every $k$ , so $(u_{k})$ converges to $0$ in $L^{2}$ and is norm bounded in $W^{1, 2} (0, 1)$ .

1.2F1algebra

The values of $I$ . Since $∣ u_{k}^{'} ∣ = 1$ almost everywhere, $I (u_{k}) = \int_{0}^{1} (u_{k}^{' 2} - 1)^{2} = 0$ for every $k$ ; and $I (0) = \int_{0}^{1} (0 - 1)^{2} d x = 1$ . Hence ${lim inf}_{k} I (u_{k}) = 0 < 1 = I (0)$ .

2.1F2F3F4step 1.1

$u_{k} ⇀ 0$ in $W^{1, 2} (0, 1)$ . Suppose not; then there are a bounded linear functional $F$ on $W^{1, 2} (0, 1)$ and $ε > 0$ with $∣ F (u_{k}) ∣ \geq ε$ for infinitely many $k$ . Along that subsequence, which stays norm bounded, [F3] provides a further subsequence with $u_{k_{l}} ⇀ y$ for some $y \in W^{1, 2} (0, 1)$ . For every $φ \in L^{2} (0, 1)$ the bounded functional of [F2] gives $\int_{0}^{1} u_{k_{l}} φ \to \int_{0}^{1} y φ$ , while $\int_{0}^{1} u_{k_{l}} φ \to 0$ because $u_{k_{l}} \to 0$ in $L^{2}$ by step 1.1; passing to the limit, $\int_{0}^{1} y φ = 0$ for all $φ \in L^{2}$ , so $y = 0$ by taking $φ = y$ in [F4]. But then $F (u_{k_{l}}) \to F (0) = 0$ , contradicting $∣ F (u_{k_{l}}) ∣ \geq ε$ . Hence $u_{k} ⇀ 0$ in $W^{1, 2} (0, 1)$ .

3.1F5step 1.2step 2.1algebra∎

Conclusion and scope. Steps 1.2 and 2.1 give $I (0) = 1 > 0 = {lim inf}_{k} I (u_{k})$ along a sequence converging weakly to $0$ , so $I$ is not weakly sequentially lower semicontinuous. Since $I (- u_{1}) = I (u_{1}) = 0$ but $I (0) = 1$ , $I$ is not convex. Yet $I \geq 0$ and $I (u_{1}) = 0$ , so its minimum is attained. Thus the example shows loss of weak lower semicontinuity for a nonconvex gradient energy, without asserting necessity of convexity for existence; the cited convex integral theorem also has dimensional and growth hypotheses absent here.

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Sources

Francesco Paolo Maiale (course by Giovanni Alberti), Lecture Notes Calculus of Variations A, University of Pisa (last update 21 August 2019; complete 149-page notes) (standard reference, not scraped)
Gerald Teschl, Partial Differential Equations: From Classical to Modern (2026 author manuscript; complete 392-page archived text) (standard reference, not scraped)